How do you solve #x^2-2x-1=x+3#?

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Answer 1 · 2018-04-13T05:15:13

#x=4#
#x=-1#

First, get #0# on one side.
#(x^2-2x-1)# #-(x+3)#=#x+3# #-(x+3#)

#x^2-3x-4=0#

Next, factor this equation.
#(x-4)(x+1)=0#
Therefore,
#(x-4)=0#
#x=4#
or
#(x+1)=0#,
#x=-1#

You can verify this using FOIL:
=#x^2+x-4x-4#
=#(x^2-3x-4)#

Answer 2 · 2018-04-13T05:22:02

#x=-1#
#x=4#

#x^2-2x-1=x+3#

#x^2-2x-1-x-3=0#

#x^2-3x-4=0#

#x^2-4x+x-4=0#

#x(x-4)+1(x-4)=0#

#(x+1)(x-4)=0#

#x+1=0#
#x=-1#

#x-4=0#
#x=4#

Answer 3 · 2018-04-13T05:22:45

#x=-1,x=4#

Consider the given equation
#x^2-2x-1=x+3#

Subtract #x+3# from both sides (Or bring all terms to the left-hand side)

#x^2-2x-1-x-3=x+3-x-3#

Thus,
#x^2-2x-1-x-3=0#

Add like terms
#x^2-3x-4=0#

Split middle term so as to obtain #(-4)# as product and #(-3)# as sum
#x^2+x-4x-4=0#

Take common factors out from paired terms
#x(x+1)-4(x+1)=0#

Then,
#(x+1)(x-4)=0#
#x+1=0# or #x-4=0#
#x=-1# or #x=4#